The equation x^3+x-1=0 has | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given equation is $f(x) = x^3+x-1 = 0$.To determine the number of real roots,we analyze the derivative of the function $f(x)$.The derivative is $f

Vedclass · Accepted Answer

exactly one real root.

Vedclass · Answer

(C) Given equation is $f(x) = x^3+x-1 = 0$.To determine the number of real roots,we analyze the derivative of the function $f(x)$.The derivative is $f'(x) = 3x^2 + 1$.Since $x^2 \geq 0$ for all $x \in \mathbb{R}$,it follows that $3x^2 + 1 \geq 1 > 0$ for all $x \in \mathbb{R}$.Because $f'(x) > 0$ for all $x$,the function $f(x)$ is strictly increasing on the entire real line.$A$ strictly increasing function can cross the $x$-axis at most once.We observe the values of the function at specific points:$f(0) = 0^3 + 0 - 1 = -1$$f(1) = 1^3 + 1 - 1 = 1$Since $f(0)  0$,by the Intermediate Value Theorem,there must exist at least one real root $x \in (0, 1)$ such that $f(x) = 0$.Since the function is strictly increasing,this root is unique.Therefore,the equation has exactly one real root.

The equation $x^3+x-1=0$ has

Similar Questions

$f(x) = x^2 - 6x + 10$ is an increasing function in the . . . . . . interval.

If the tangent drawn to the curve $y=x^3-ax^2+x+1$ at each point $x \in R$ is inclined at an acute angle with the positive direction of the $X$-axis,then the set of all possible values of '$a$' is

$f(x) = x^3 - 27x + 5$ is an increasing function,when

If $f(x) = \int\limits_0^x {{e^t}{{\sin }^{ - 1}}(t - 1)\ln t\,dt}$ for $x > 0$,then:

In which interval is the function $f(x) = 2x^2 - \ln |x|$ $(x \ne 0)$ monotonically decreasing?

Vedclass Test Series

Exam Paper Generator

Online Exam Module